Before looking at these solutions, please work the practice quiz.
Disclaimer: This quiz is representative of the level of difficulty you should expect, but it doesn’t include every single topic from the week’s work. The real quiz may include some other topics and may skip some that are in this practice quiz. (The real quiz also may not word questions in the same way as the practice quiz. You should focus on the concepts, not a particular form of words.)
See also: How to Take a Math Test
These solutions show about the same level of work I expect from you, though I often add some extra commentary. Please see Show Your Work for the what, why, and how.
1.6 1.4 1.6 0.9 1.3 1.4 1.5 1.4 1.2 1.3 1.7
Estimate the average salt content of all chips, by constructing a 95% confidence interval in endpoint form. Interpret your answer in English.
Solution: This is numeric data, one population mean with population standard deviation unknown: Case 1 on Inferential Statistics: Basic Cases.
Make sure the requirements are satisfied. Since the sample size is under 30, you must check that the sample data are normally distributed with no outliers. Your two test screens should look like this:
You don’t need to copy the screens, but you should say that you used MATH200A part 2 and found no outliers, and you used MATH200A part 4 and found a normal distribution (r=.9624, crit=.923, r>crit).
Since you don’t know the standard deviation of the population, use a T-Interval to compute the confidence interval.
And you show your work as
TInterval inputs: List:L1, Freq:1, C-Level:.95
outputs: (1.2423,1.5395), x̅=1.39; s=0.22; n=11
Answer: We have 95% confidence that the average salt content of all chips is between 1.24 and 1.54 percent.
Remark: By convention, we round means and standard deviations to one more decimal place than the original data. It’s not the end of the world if you do it differently, but following the convention is usually a good choice.
Common mistake: Make sure that your confidence interval refers to the population, not to the sample. Any reference to “11 batches” in your confidence interval is wrong. You’re not estimating anything about the 11 batches; you know their mean exactly.
If you prefer, you can do
1-Var Stats before
TInterval. If you do that, the calculator pastes the
values into the TInterval screen for you:
The output screen would be the same, of course. You show your work as
TInterval x̅:1.3909, s:0.2212, n:11, C-Level:.95
Answer: p̂ = 5067/10000 = 0.5067
Remark: Students seem to have an awful lot of trouble with the term “point estimate”. Don’t make it harder than it needs to be: the point estimate for the population mean (or proportion, standard deviation, etc.) is very simply the sample mean (or proportion, standard deviation, etc.).
Solution: This is Case 2, one population proportion, a sample size problem. You have a prior estimate of 42% or 0.42 for the true proportion. Use MATH200A part 5. Enter
Data Type: 3:Binomial, p̂: .42, E: .035, C-Level: .9
and read off the answer: n ≥ 539
Common mistake: Convert percentages to decimals carefully! 3.5% is 0.035, not 0.35 or 3.5.
Common mistake: Always use a prior estimate when you have one. If you use the conservative figure of p̂ = 0.5, you get a sample size of 553, which is larger than necessary.
Alternative solution: If you don’t have the program, or if you prefer to work out the solution, use the formula at right. You have a prior estimate of 42% or 0.42 for the true proportion, so you use p̂ = 0.42 and 1−p̂ = 0.58. E is 3.5% or 0.035 (given).
For z(α/2), start with 1−α = 0.90; therefore α = 0.10 and α/2 = 0.05. Now you need z0.05, which is invNorm(1−0.05) or about 1.6449. (From Chapter 7, z0.05 is the z score that divides the normal curve so that the area to left is 0.05.) Substituting,
n = .42 * (1−.42) * (invNorm(1−0.05)/0.035)² = 538.0166
Since the sample size must always be a whole number, greater than or equal to the computed n, your sample must be at least 539 people.
See also: How Big a Sample Do I Need?
Common mistake: When you get the answer of 538-point-something, don’t stop there. Sample sizes are always rounded up to the next whole number.
Solution: This is Case 0, numeric data with known s.d. of the population. Use MATH200A part 5 again. Enter
Data Type: 1:Num known σ, σ: 600, E: 250, C-level: .98
and the answer is n ≥ 32
Alternative solution: If you don’t have the program, use the formula at right. Here σ = 600, E = 250, and you have to find zα/2. (From Chapter 7: zα/2 is the z score that divides the normal curve so that the area to left is α/2.)
C-Level = 1−α = .98 ⇒ α = .02 ⇒ α/2 = .01
zα/2 = z.01 = invNorm(1−.01) ≈ 2.3263
n = ( invNorm(1−.01) * 600 / 250 )² = 31.1725 → 32 people
Common mistake: The answer is not 31 people because you don’t round sample size. Rather, you interpret it as “I need at least 31.1725 responses”, which means 31 won’t be enough but 32 is adequate.
Solution: This is binomial data for one population: Case 2.
1-PropZInt: x=61, n=100, C-Level=.95
Outputs: (.5144,.7056); p̂=.61
Check the requirements, using p̂ from the calculator.
Is np̂(1−p̂)≥10? p̂ = 0.61 and n = 100. np̂(1−p̂) = 100*.61*(1−.61) = 23.79; check.
Is 20n≤N? 20*100 = 2000, and the TC3 student population numbers greater than 2000; check.
Answer: We’re 95% confident that 51.4% to 70.6% of all TC3 students believe in angels.
Common mistake: Again, your statement needs to be about the population (all TC3 students), not about the sample (100 students). There’s no estimation related to the 100 students, because you know that exactly 61 of them believe in angels.
Remark: It’s possible that two thirds of TC3 students believe in angels. We can't really tell because 66.7% is in the 95% confidence interval.